SERIES—CUBES AND OTHERS 33

Since both [100n%—2(10)7] and [10072+2(10)7] must
clearly be multiples of 10, it follows that both squares must
be one more than the multiples of 10. This is another way of
saying that they will both have 1 as end-digit.

For the same reason, the squares of any two numbers hav-
ing end-digits which are complementary (i.e. add together to
make 10; such as 8 and 2, 7 and 3, etc.) will themselves have
identical end-digits.

The same principle can be applied to cube numbers. Any
cube ending in g can be written (10n—1)2% and this equals
(ton—r1)[(10m)2—2(10)n+1]. If the brackets are removed
all the terms in the final result, except the last term, will be
multiples of 10. The last term is (—1)(+1)= —1. The ex-
pression is therefore a multiple of 10, less 1; and the end-digit
is therefore g.

The end-digits of squares are limited to o, 1, 4, 5, 6 and q.
Those of cubes, however, use up o and all nine digits. Num-
bers ending in o, 1, 4, 5, 6 or g have cubes with the same end-
digits respectively, whereas numbers ending in 2, 3, 7 or 8
have cubes the end-digits of which are complementary to 10.
Thus if the end-digit of a number is 2, then its cube will have
the end-digit 8.

As fourth powers are the squares of square numbers it
follows that the end-digits of fourth powers are limited to the
end-digits of the squares of o, 1, 4, 5, 6 and g—thatis o, 1, 6
or 5. Fifth powers have exactly the same end-digits as have
their roots. This is another way of saying that the difference
between any two consecutive fifth power numbers is always
a number having 1 as its end-digit.

28 —1%= g2 . 1= g1
6°—5°=7776 —3125=4651
This can be proved as follows:
(n+1)8 =n®+5nt+ 100+ 541
(n41)8—nd=s5nt+ 1003+ 1002450 41
=5(n*4-2n3+2n%+4n) 41
But (n*+2n®-+2n2+n) must be even and can be represented
3

O R BT P

R R

SRR A AR

MEAHEBRAA AR

T T AR N AR nT

k N A RS TR T N T AR B A e O AR T
W»«.fiwwmwdfimmmxwwm SRR RHEARL

 

!